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In the following example verify that the given expression is a solution of the corresponding differential equation: y = eaxxdydxyyeax;xdydx=ylogy

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Question

In the following example verify that the given expression is a solution of the corresponding differential equation:

y = `"e"^"ax"; "x" "dy"/"dx" = "y" log "y"`

Sum
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Solution

y = `"e"^"ax"`

∴ log y = log `"e"^"ax"` = ax log e

∴ log y = ax        .....(1) .....[∵ log e = 1]

Differentiating w.r.t. x, we get

`1/"y" * "dy"/"dx" = "a" xx 1`

∴ `"dy"/"dx" = "ay"`

∴ `"x""dy"/"dx" = ("ax")"y"`

∴ `"x" "dy"/"dx" = "y" log "y"`     ....[By (1)]

Hence,  y = `"e"^"ax"` is a solution of the D.E. `"x" "dy"/"dx" = "y" log "y"`.

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Chapter 6: Differential Equations - Exercise 6.3 [Page 200]

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